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Unformatted text preview: Lecture 13 In (15) of Lecture 11, the electrical equations of an electric machine are put in the first order standard form: ) , ( u x f dt x d = (1) where u is a vector of the voltages and the load torque T L , ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = m i x ω and where ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = δ γ δ γ R R S S i i i i i . Specifically in the γ δ frame, the equations are: } ] [ ] [ ] [ { ] [ 1 i F i G n i R e L i p m − − − = − ω (2) 1 1 / )} ( { J i i i i nM T f p R S R S L m m γ δ δ γ ω ω − + − − = (3) In general, the equations are solved by numerical integration. It is important to understand the concept the “state” ) ( t x at time t in “statespace”. Specifying the state ) ( t x at time t means that it is not necessary to know the history of ) ( t x for t t ≤ ≤ ∞ − . On the other hand, the initial condition ) ( t x at time t determines the solution of ) , ( u x f dt x d = for t t ≤ . The integration software requires the initial state because without it, it cannot evaluate ) ), ( ( u t x f dt x d = , the initial gradient to start the integration. SteadyState Solutions γ δ frame In the γ δ frame, because u is time invariant, one expects that X t x = ) ( is also time invariant. As such, the steadysolution satisfies = dt x d , which means that ) , ( = u x f . This can be reached numerically integrating (2) and (3) long enough until a time invariant solution X t x = ) ( is reached....
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This note was uploaded on 02/23/2010 for the course ECSE 462 taught by Professor Bontekooi during the Spring '09 term at McGill.
 Spring '09
 BonTekOoi

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